7
$\begingroup$

Let x be a positive element in the spatial tensor product of two non unital C* algebras A and B. Is there a single element $a \otimes b \geq x$? How can we noncommutativize the following proof, in the commutative case: Let $F^2$ be a positive function on $X\times Y$. Define $f(x)=\sup_{y\in Y} F(x,y)$ and $g(y)=\sup_{x\in X} F(x,y)$, then $F^2 \leq fg$.

$\endgroup$
12
$\begingroup$

Yes. For a self-adjoint element $y$, denote $s(y)=\sup{\rm Sp}(y)$. Then for $\gamma \geq s(y)$, one has $$\inf \lbrace s(y - \gamma(e\otimes f)) : 0\le e\le 1,\ 0\le f\le 1\rbrace \le 0.$$ Indeed, if $e_n$ and $f_n$ are approximate units, then so is $g_n:=e_n\otimes f_n$ and $y - \gamma g_n \le y-g_n^{1/2} y g_n^{1/2} \to 0$. Now let $y_0 := x \le 1$ and find $0 \le e_n \le 1$ and $0 \le f_n \le 1$ recursively so that the elements $y_{n+1} := y_n - 4^{-n}(e_n \otimes f_n)$ satisfy $s(y_{n}) \le 4^{-n}$ for all $n$. Let $e = \sum_{n=0}^\infty 2^{-n}e_n$ and likewise for $f$. Then, one has $$x = y_0 \le \sum_n 4^{-n}e_n\otimes f_n \le e\otimes f.$$ If $A$ and $B$ have strictly positive elements, one can arrange $(e\otimes f) - x$ is strictly positive.

I think with more efforts one can find $e$ and $f$ such that $\| e \| \| f \| = \| x \|$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you, But could you please explain why the last inequality hold? $\endgroup$ – Ali Taghavi Jul 11 '13 at 11:03
  • $\begingroup$ That's because $e \otimes f = \sum_{m,n}2^{-(m+n)}e_m \otimes f_n$ is an unconditionally norm convergent series of positive elements. $\endgroup$ – Narutaka OZAWA Jul 11 '13 at 23:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.